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我正在爲我的課程之一使用旋轉式手機和arduino套件創建基於自動手機菜單的遊戲進行體驗設計項目。旋轉撥號的串行輸入通過arduino運行，現在我正在使用處理來編寫菜單。如何編寫嵌套if或case來提示輸入菜單？
我有一個行動綱要，並已開始編碼一些，如果然後語句去，但現在我偶然發現案件和切換。
我對這個完全陌生，但在課堂上學到了很多東西。
我的問題是如何製作嵌套的if/then命令的連續集或使用大小寫和切換來移動一系列提示和輸入？
這裏是我的草圖至今：
import processing.serial.*;
Serial port; // Create object from Serial class
float val; // Data received from the serial port
boolean task1prompted;
boolean task1;
boolean task2;
boolean dialed;
PFont font;
void setup() {
size(800, 400);
background(0, 0, 0);
smooth();
// IMPORTANT NOTE:
// The first serial port retrieved by Serial.list()
// should be your Arduino. If not, uncomment the next
// line by deleting the // before it. Run the sketch
// again to see a list of serial ports. Then, change
// the 0 in between [ and ] to the number of the port
// that your Arduino is connected to.
//println(Serial.list());
String arduinoPort = Serial.list()[0];
port = new Serial(this, arduinoPort, 9600);
task1 = false;
task2 = false;
task1prompted = false;
font = createFont("Arial", 32);
textFont(font, 32);
textAlign(CENTER);
}
void draw() {
if (port.available() > 0) { // If data is available,
val = port.read(); // read it and store it in val
if (val >= 48 && val <= 57) {
val = map(val, 48, 57, 0, 9); // Convert the value
}
println(val);
}
if (val == 97) {
println("dialing");
}
if (val == 98){
println("dialed");
dialed = true;
}
/// switch will activate the task1 variable.
// Play sound file for the prompt.
if (task1prompted == false){
delay(1000);
println("for spanish, press one. for french, press 2...");
task1prompted = true;
}
task1 = true;
if (task1 == true && dialed == true) {
///play sound file
if (val == 5) {
println("Thank you for playing... Blah blah next prompt.");
dialed = true;
task1=false;
task2=true;
} else
if (val != 5) {
println("We're sorry, all of our international operators are busy");
task1 = true;
task2 = false;
dialed = false;
}
}
else
if (task2 == true){
delay(1000);
println("task2 start");
}
}
我的教練幫助我走到這一步，我已經沖刷有關如何繼續前進到下一個任務/提示答案。使用案例和切換會更容易嗎？我甚至在做嵌套的if語句是否正確？
好吧，我只是嘗試了這一點與草圖和案例命令如下：
/// Switch will activate the task1 variable.
// Play sound file for the prompt.
if (task1prompted == false){
delay(1000);
println("for spanish, press one. for french, press 2...");
task1prompted = true;
}
task1 = true;
if (task1 == true && dialed == true) {
///Play sound file
int lang = (int)(val+0);
switch(lang) {
case 1:
case 2:
case 3:
case 4:
println("sorry no international operators"); // If 1-4 go back to choices
task1 = true;
task2 = false;
dialed = false;
break;
case 5:
println("thank you, move to next prompt"); // If 5 go to next prompt
task1=false;
task2=true;
dialed = true;
break;
case 6:
case 7:
case 8:
case 9:
case 0:
println("not a valid option, you lose"); // If not 1-5 go back to beginning
task1=false;
task2=false;
dialed = true;
break;
}
if (task2prompted == false){
delay(1000);
println("please listen while we test the line");
task2prompted = true;
}
task2 = true;
if (task2 == true && dialed == true) {
} ///Play sound file
int tone = (int)(val+0);
switch(tone) {
case 1:
case 2:
case 3:
case 5:
case 6:
case 7:
case 8:
case 9:
case 0:
println("not a valid connection, you lose"); // If not 4 go back to beginning
task2 = false;
task3 = false;
dialed = false;
break;
case 4:
println("thank you, move to next prompt"); // If 4 move to next prompt
task2=false;
task3=true;
dialed = true;
break;
}
}
}
我還在困惑如何使這個有水平，不是所有的同時發生。